S
93102
931020
For Supervisor’s use only
Scholarship 2008
Chemistry
9.30 am Saturday 15 November 2008
Time allowed: Three hours
Total marks: 40
Check that the National Student Number (NSN) on your admission slip is the same as the number at the
top of this page.
You should answer all the questions in this booklet.
A periodic table is provided on page 2 of this booklet.
Write all your answers in this booklet.
If you need more space for any answer, use the page(s) provided at the back of this booklet and clearly
number the question.
Check that this booklet has pages 2–22 in the correct order and that none of these pages is blank.
YOU MUST HAND THIS BOOKLET TO THE SUPERVISOR AT THE END OF THE EXAMINATION.
© New Zealand Qualifications Authority, 2008
All rights reserved. No part of this publication may be reproduced by any means without the prior permission of the New Zealand Qualifications Authority.
Be
9.0
K
Rb
Cs
Scholarship Chemistry 93102, 2008
Fr
Ca
Sr
Ba
Ra
Actinide
Series
Y
Lu
La
Ac
Lr
Zr
Hf
Ce
Th
Rf
Nb
Ta
Pr
Pa
Db
W
Nd
U
Sg
Mn
7
Tc
Re
237
Np
Bh
Ru
Os
239
Pu
Hs
Rh
Ir
Eu
Mt
241
Pd
Pt
244
Cm
Bk
Au
Cf
Hg
Ga
Tl
Ge
Pb
257
Fm
P
Sb
Bi
Tm
258
Md
S
259
No
Po
210
84
Yb
Te
128
52
79.0
Se
32.1
34
102
O
16
16.0
16
8
173
70
209
83
122
51
74.9
As
31.0
33
101
N
15
14.0
15
7
169
69
207
82
119
Sn
72.6
50
100
Si
28.1
32
Er
C
14
12.0
14
6
167
68
204
81
Ho
In
115
49
69.7
31
Es
Al
27.0
252
99
B
13
10.8
13
5
165
67
201
80
Dy
Cd
112
48
251
98
Zn
12
65.4
30
163
66
197
79
Tb
Ag
108
47
249
97
Cu
11
63.6
29
159
65
195
78
106
46
Gd
96
Ni
10
58.7
28
157
64
268
109
192
77
103
45
Am
95
Co
9
58.9
27
152
63
265
108
190
76
101
44
Sm
94
Fe
55.9
26
8
Molar Mass / g mol–1
150
62
264
107
186
75
98.9
43
54.9
Pm
93
H
1.0
25
1
147
61
263
106
184
74
95.9
Mo
42
238
92
Cr
6
52.0
24
144
60
262
105
181
73
92.9
41
231
91
V
5
50.9
23
141
59
261
104
179
72
91.2
40
232
90
Ti
4
47.9
22
140
58
262
103
175
71
88.9
39
227
89
Sc
3
45.0
21
139
57
226
88
137
56
87.6
38
40.1
20
Lanthanide
Series
223
87
133
55
85.5
37
39.1
19
Mg
24.3
12
4
Na
6.9
Li
2
23.0
11
3
1
Atomic Number
PERIODIC TABLE OF THE ELEMENTS
F
Br
I
At
210
85
127
53
79.9
35
35.5
Cl
19.0
17
9
17
He
Ar
Kr
Xe
Rn
222
86
131
54
83.8
36
40.0
18
20.2
Ne
4.0
10
2
18
2
3
You have three hours to complete this examination.
Assessor’s
use only
QUESTION ONE (8 marks)
(a)
Labels have fallen off six bottles known to contain the following aqueous solutions.
NaCl
KI
Na2CO3
Fe(NO3)3
NaOH
Zn(NO3)2
Outline a procedure that uses only the contents of the unlabelled bottles to identify each
solution. Include balanced equations for any reactions that occur.
Use the following information where it is relevant.
E°(Cl2 / Cl–) = +1.36 V
E°(I2 / I–) = +0.54 V
E°(Fe3+ / Fe2+) = +0.77 V
E°(Fe2+ / Fe) = –0.44 V
Scholarship Chemistry 93102, 2008
E°(Zn2+ / Zn) = –0.76 V
4
(b) A sample of water is to be analysed to determine the chloride ion content. Before carrying
out the analysis, a pH meter is used to ensure the pH of the water is in the range 7–8. If the
solution is too basic, 1 mol L–1 ethanoic acid is added dropwise until the pH is reduced to just
below 8. If the solution is too acidic, 1 mol L–1 sodium carbonate solution is added until the
pH is increased to just below 8.
A 20.00 mL sample of water containing chloride ions was placed in a conical flask and
3.00 mL of 0.100 mol L–1 K2CrO4 solution was added as an indicator. The resulting solution
was then titrated with 0.0500 mol L–1 silver nitrate solution, causing the formation of a
precipitate of AgCl. 18.35 mL was required to reach the end-point of the titration, the point
when there is a slight red-brown tinge due to the formation of a red-brown precipitate of silver
chromate.
(i)
In terms of the species present in the flask during the titration, explain why there would
be a problem in the analysis if the solution were very acidic, and also why there would
be a problem if the solution were very basic.
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(ii) Calculate the concentration of chloride ions remaining in the solution at the point
when the silver chromate starts to form. Show that the amount, in moles, of chloride
ions remaining in the solution at the end-point is less than 0.1% of the original amount
present.
Ks (AgCl) = 1.77 × 10–10
Ks (Ag2CrO4) = 2.60 × 10–12
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QUESTION TWO (8 marks)
(a)
In order to identify the organic compound A with molecular formula C10H18O3, the following
series of reactions was carried out.
Assessor’s
use only
A
dil. H2SO4
B
conc. H2SO4
D
MnO4– / H+
E
E
Cr2O72– / H+
F: C4H6O5
F
conc. H2SO4
G: C4H4O4
B: C4H8O3 + C: C6H12O
D: C4H6O2
Deduce the structures of Compounds A to G.
Additional information:
Compound A exists as enantiomers (optical isomers).
Compound C has a six-membered ring.
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Assessor’s
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8
(b) The compound X can be synthesised from the given starting materials: benzoic acid,
methanamine and 4–chloro–3,5–dimethylcyclopent–1–ene.
O
C
H3C
O
N
CH3
O
C
CH3
OH
H3C
NH2
Compound X
Starting Materials
Note: For clarity, the hydrogens have been left off the cyclopentane and benzene rings.
The following reagents are available:
Draw a reaction scheme to show how compound X can be synthesised from the starting
materials. Give an explanation for the order in which the reactions are carried out.
Cl
CH3
CH3
Assessor’s
use only
SOCl2
NaOH(aq)
NaOH(in ethanol)
Conc HCl
MnO4– / H+
Cr2O72– / H+
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QUESTION THREE (8 marks)
(a)
Assessor’s
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Nitrogen trihalide compounds have varying stability. At room temperature, nitrogen triiodide,
NI3, is a black-red solid. When it is touched, it decomposes violently into its elements,
nitrogen gas and solid iodine. At room temperature, nitrogen trichloride, NCl3, is a pale
yellow oil that is explosive. In contrast, nitrogen trifluoride, NF3, is a colourless gas, which
does not react with water, and only reacts with most metals on heating. Unlike NH3, NF3 does
not act as a base.
DATA
Average bond enthalpies (kJ mol–1)
Pauling electronegativities
F–F
+159
H
2.20
I–I
+151
N
3.04
N–F
+278
F
3.98
N–I
+159
Cl
3.16
N≡N
+945
I
2.66
ΔsubH°(I2) = 62 kJ mol–1
Account for:
(i)
(ii) the different stability of NF3 and NI3 towards decomposition
(iii) the differing basicities of NF3 and NH3.
the different states of the nitrogen trihalide compounds at room temperature
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(b) (i)
A lithium atom has 3 electrons. The accepted model for the ground state electronic
structure of Li is that 2 electrons occupy the lowest energy level, with the remaining
electron occupying a higher energy level.
1st IE / kJ mol–1
Li
526
He
2379
Discuss how the relative ionisation energies of Li and He are consistent with this model
rather than the alternative that has all 3 electrons of a lithium atom in the lowest energy
level.
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(ii) Match the following atoms and ions to the radii given. Justify your answer.
Atom or Ion
Ga
Ga3+
Se2–
Se
S
Radius / pm
122
198
62
103
135
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QUESTION FOUR (8 marks)
(a)
(i)
Assessor’s
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A ‘Latimer Diagram’ can be used to summarise the reduction potential data for elements
that exist in several oxidation states in aqueous solutions. The oxidation numbers
decrease from left to right and E° for each couple is written above the line joining the
two species involved in the couple.
The Latimer Diagram below shows the standard electrode potentials for chlorine in
basic conditions.
ClO4– +0.37 V ClO3– +0.30 V ClO2– +0.68 V ClO– +0.42 V Cl2 +1.36 V Cl–
In disproportionation reactions, an atom in a particular oxidation state acts as both the
oxidant and the reductant.
Account for the disproportionation of Cl2 in basic aqueous solution, and identify any
other chlorine species that will disproportionate in these conditions. Explain how the
extent of disproportionation of Cl2 is pH dependent.
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(ii) At a pH of 6.5, an aqueous solution of hypochlorous acid, HClO, at 20°C is 7.6%
dissociated.
Calculate the pKa of HClO and hence find the pH of a 40 mg L–1 solution of Ca(ClO)2.
M(Ca(ClO)2) = 143.0 g mol–1
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(b) Solutions of quinine, like those of sodium hypochlorite, NaClO, are basic. Solutions of
quinine can be standardised by titration with hydrochloric acid.
(i)
Explain why hydrochloric acid is not used to standardise NaClO.
(ii) The solubility of quinine is 0.577 g L–1.
pKa (quinineH+) = 8.9
M(quinine) = 324.4 g mol–1
Calculate the pH at the equivalence point when 100.0 mL of a saturated solution of
quinine is titrated against 0.0100 mol L–1 HCl.
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QUESTION FIVE (8 marks)
(a)
Assessor’s
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A calorimeter that is perfectly heat insulated is filled with water that is initially at a
temperature of 22.55°C.
7.80 g of solid ZnSO4 is added, and the mixture is stirred until the solid completely dissolves.
The temperature of the water increases to 23.52°C.
In a second experiment the same calorimeter is filled with water that is initially at a
temperature of 22.15°C. When 12.30 g of the salt ZnSO4.7H2O is added to the water the
temperature drops to 21.84°C as a result of the dissolution process.
The heat capacity of the calorimeter and its contents is 0.900 kJ °C–1.
Calculate ∆rH for the following process:
ZnSO4(s)
+
7H2O(ℓ) → ZnSO4.7H2O(s)
M(ZnSO4) = 161.5 g mol–1
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(b) The box below contains data relating to the dissolving process for MgCl2 and AgCl.
The lattice enthalpy, ΔLEH°, of an ionic solid is the enthalpy change when 1 mole
of the ionic solid is separated into its ions in the gas phase. For example the lattice
enthalpy for MgCl2 is the energy change for the reaction:
MgCl2(s) → Mg2+(g) + 2Cl–(g)
The enthalpy of solution is the enthalpy change that occurs when 1 mole of solid
dissolves to form a solution.
The enthalpy of hydration, ΔhydH°, is the enthalpy change when 1 mole of gaseous
ions is hydrated. For example:
Mg2+(g) → Mg2+(aq)
Melting points
/ °C
Lattice enthalpy
ΔLEH° / kJ mol–1
MgCl2
714
MgCl2(s)
2523
AgCl
455
AgCl(s)
915
ΔhydH°
/ kJ mol–1
Ag+
–474
Mg2+
–1931
Cl–
–361
Explain why it is possible to dissolve some ionic solids in water in spite of the strong ion–ion
forces of attraction in the crystal lattice of the solid. Use the data given to support your
explanation and to outline factors that contribute to MgCl2 being a more soluble salt than
AgCl.
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Clearly number the question.
Question
number
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Clearly number the question.
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number
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number
Scholarship Chemistry 93102, 2008
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93102
For Assessor’s use only.
Keep flap folded in.
For Assessor’s Use Only
Question Number
Marks
ONE
(8)
TWO
(8)
THREE
(8)
FOUR
(8)
FIVE
(8)
TOTAL
(40)
Scholarship Chemistry 93102, 2008